If $ax^2 + 2hxy + by^2 =0$ be the two sides of a parallelogram and $px + qy = 1$ be its one diagonal then prove that the equation of the other diagonal is $y(bp-hq)=x(aq-hp)$.
While searching for the answers, I got this answer on 'Yahoo answers' but I did not understand anything over here. Please anyone make me clear about it.
Let the parallelogram be $OPMQ$ ($PQ$ given diagonal) then $$ax²+2hxy+by²+λ(px+qy−1)=0$$ is a conic through $P$ and $Q$. For this to be 2nd pair of sides of parallelogram it must be of form $$a(x−X)²+2h(x−X)(y−Y)+b (y−Y)²=0$$ ( that is the same as given pair of sides but translated to $M(X,Y)$ as origin ) Quadratic terms are of course identical. Comparing linear terms gives $$−2aX −2hY=λp, −2bY−2hX=λq.$$ Divide to eliminate $λ$ and get $$Y/X = (aq −hp)/(bp−hq).$$ 2nd diagonal through $O$ is $y/x=Y/X$.
Other than this answer:
Let $y = m_1x$ and $y=m_2x$ be the two lines represented by $ax^2+2hxy+by^2=0$ so that $$m_1+m_2 = -\frac{2h}{b} \text{ and } m_1m_2=\frac{a}{b}...(1)$$ Let $OA$ and $OC$ represent these lines and let the line $px+qy=1$ meets $OA$ and $OC$ at $A$ and $C$. After completing te parallelogram, we have to find the equation of the diagonal $OB$. Let the two diagonals intersect at $M(x_1, y_1)$.
The line $px+qy=1$ meets $y=m_1x$ at $A(\frac{1}{p+qm_1}, \frac{m_1}{p+qm_1})$ and $y=m_2x$ at $C(\frac{1}{p+qm_2}, \frac{m_2}{p+qm_2})$. Now the coordinates of the midpoint of $A$ and $C$ are given by $$M(x_1, y_1) = M(\frac{2p+q(m_1m_2)}{2(p+qm_1)(p+qm_2)}, \frac{p(m_1+m_2)+2qm_1m_2}{2(p+qm_1)(p+qm_2)})...(2)$$ Equation of the other diagonal $OB$ is thus, $$y-0 = \frac{y_1-0}{x_1-0}(x-0) \Rightarrow \frac{y}{x} = \frac{y_1}{x_1}$$ Substituting the coordinates from $(2)$ gives us the required answer.